The usual second isomorphism theorem for groups is: let $G$ be a group, $S$ and $N$ subgroups with $N$ normal, then $SN$ is a subgroup of $G$, $S\cap N$ is a normal subgroup of $S$ and $SN/N \simeq S /S \cap N$.
($N$ need not to be a normal subgroup as long as $S$ is a subgroup of the normalizer of $N$).

Now if $S$ and $N$ are normal subgroups of $G$ then $SN/N \simeq S /S \cap N$ and $SN/S \simeq N /S \cap N$, so that the normal chains $(S \cap N \subset S \subset SN)$ and $(S \cap N \subset N \subset SN)$ are equivalent.

It's this last property that we would like to generalize to the inclusions of groups $(H \subset G)$.

The notion of normal subgroup $K \triangleleft G$ (i.e. $ \forall g \in G \text{ , } Kg=gK $) can be generalized by the notion of normal intermediate subgroup (motivated by the prop.3.3 p476 of this paper).

Definition: $K$ is a normal intermediate subgroup of the inclusion $(H \subset G)$ if $H \subset K \subset G$, and $$\forall g \in G \text{ , } KgH=HgK $$ Remark: If $K_i$ is a normal intermediate subgroup of $(H \subset G)$, then so are $K_1 \cap K_2$ and $K_1K_2$,
and if $H \subset H_0 \subset K_i \subset G_0 \subset G$ then $K_i$ is also a normal intermediate subgroup of $(H_0 \subset G_0)$.

Definition: A chain of subgroups $(K_1 \subset K_2 \subset \dots \subset K_n)$ is a normal intermediate chain if $K_i$ is a normal intermediate subgroup of the inclusion $(K_{i-1} \subset K_{i+1})$.

Definition: Two chains of subgroups $(K_1 \subset K_2 \subset \dots \subset K_n)$ and $(L_1 \subset L_2 \subset \dots \subset L_m)$ are equivalent if $m=n$ and $\exists \sigma \in S_{n-1}$ such that $(K_i \subset K_{i+1}) \sim (L_{\sigma(i)} \subset L_{\sigma(i)+1})$.

Definition: Two inclusions of groups are equivalent, $(A \subset B) \sim (C \subset D)$, if: $(A/A_B \subset B/A_B) \simeq (C/C_D \subset D/C_D)$ with $A_B$ the normal core of $A$ in $B$.

Question: Let $K$ and $L$ be normal intermediate subgroups of an inclusion $(H \subset G)$ then, are the normal intermediate chains $(K \cap L \subset K \subset KL)$ and $(K \cap L \subset L \subset KL)$ equivalent ?

Examples: See this post and note that this answer gives here an example.

Motivation: A Jordan-Hölder theorem generalized to the inclusions of groups.


1 Answer 1


No, the following theorem gives counterexamples:

Theorem: $\forall n \ge 3$, $A_{n+1}$ and $S_n$ are normal intermediate subgroups of the inclusion $(A_n \subset S_{n+1})$, but $(A_n \subset S_n \subset S_{n+1} )$ and $(A_n \subset A_{n+1} \subset S_{n+1} )$ are not equivalent.

Lemma: $\forall n \ge 3$, $S_{n}$ is a normal intermediate subgroup of the inclusion $(A_n \subset S_{n+1})$.

Proof: The equality $A_n . g . S_n = S_n . g . A_n$ is clear if $g \in S_n$, but $S_{n+1} = \langle S_n , \tau \rangle$ with $\tau=(n,n+1)$, so it suffices to prove that $A_n . \tau . S_n = S_n . \tau . A_n$, because $A_{n} \triangleleft S_{n}$:
If $\sigma$, $\sigma' \in S_n$ then $a_1\sigma\tau\sigma's_1=\sigma a_2 \tau s_2 \sigma=\sigma s_3 \tau a_3 \sigma = s_4\sigma\tau\sigma' a_4$ ($a_i \in A_n$, $s_i \in S_n$).

Now $S_n= \langle (1,2, \dots , n) , (1,2)\rangle$, and $A_n . (n,n+1) . (1,2) = (1,2) . (n,n+1) . A_n$
for all $n \ge 3$, so it suffices to show that $A_n . (n,n+1) . (1, \dots , n) \subset S_n . (n,n+1) . A_n$.

But $A_n . (n,n+1) . (1, \dots , n) = (n,n+1) . (1, \dots , n) . A_n$,
and $(n,n+1) . (1, \dots , n) = e.(n,n+1) . \sigma_1 = (1,2) . (n,n+1) . \sigma_2$,
with $e, (1,2) \in S_n$ and $\sigma_1 = (1, \dots , n) \in A_n$ if $n$ odd, $\sigma_2 = (1,2). \sigma_1 \in A_n$ if $n$ even.

(Idem, $(1, \dots , n) . (n,n+1) . A_n \subset A_n . (n,n+1) . S_n $)
So, $A_n . (n,n+1) . S_n = S_n . (n,n+1) . A_n $ and then $A_n . g . S_n = S_n . g . A_n$ $\forall g \in S_{n+1}$. $\square$

Proof of the theorem: $A_{n+1}$ is obviously a normal intermediate subgroup because $A_{n+1} \triangleleft S_{n+1}$.
$S_n$ is also a normal intermediate by the previous lemma. Next $(A_n \subset A_{n+1}) \not\sim (S_n \subset S_{n+1})$. $\square$

Remark: For correctness (with the question) note that $A_{n+1} \cap S_n = A_n$ and $A_{n+1}.S_n = S_{n+1}$.


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